What is (\partial)/(\partial x)((1/x+1/y)^{-1}) ?

The answer to (\partial)/(\partial x)((1/x+1/y)^{-1}) is (y^2)/((y+x)^2)

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(\partial)/(\partial x)((1/x+1/y)^{-1})

\frac{\partial}{\partial x} ((\frac{1}{x} + \frac{1}{y})^{- 1})

\frac{y ^{2}}{( y + x ) ^{2}}

Solution steps

\frac{\partial}{\partial x} ((\frac{1}{x} + \frac{1}{y})^{- 1})

Treat

y

as a constant

Apply the chain rule:

- \frac{1}{( \frac{1}{x} + \frac{1}{y} ) ^{2}} \frac{\partial}{\partial x} (\frac{1}{x} + \frac{1}{y})

= - \frac{1}{( \frac{1}{x} + \frac{1}{y} ) ^{2}} \frac{\partial}{\partial x} (\frac{1}{x} + \frac{1}{y})

\frac{\partial}{\partial x} (\frac{1}{x} + \frac{1}{y}) = - \frac{1}{x ^{2}}

= - \frac{1}{( \frac{1}{x} + \frac{1}{y} ) ^{2}} (- \frac{1}{x ^{2}})

Simplify

- \frac{1}{( \frac{1}{x} + \frac{1}{y} ) ^{2}} (- \frac{1}{x ^{2}}) : \frac{y ^{2}}{( y + x ) ^{2}}

= \frac{y ^{2}}{( y + x ) ^{2}}

cos (x) \frac{d y}{d x} + sin (x) y = 1

\frac{d}{d x} (ln (\frac{1}{x + 1}))

\int x^{3} x^{2} + 43 d x

ln (x e^{x} + 8)

f (x) = 4 (100)^{7}

What is (\partial)/(\partial x)((1/x+1/y)^{-1}) ?
The answer to (\partial)/(\partial x)((1/x+1/y)^{-1}) is (y^2)/((y+x)^2)